2 A topological interlude 2.1 Topological spaces Recall that a topological space is a set X with a topology: a collection of subsets of X, known as open sets, such that and X are open, and finiteT intersections and arbitrary unions of open sets are∅ open. We call a set F X closed if its complement X F is open. ⊆ An open cover C of X is\ a collection of open subsets of X whose union is X. A finite subcover of C is a finite subcollection whose union still con- tains X. To say that X is compact means that every open cover of X has a finite subcover. Similarly, if Z X then an open cover C of Z is a collection of open subsets whose union contains⊆ Z. We say that Z is compact if every open cover of Z has a finite subcover. If X is compact, then it is easy to show that any closed subset of X is also compact. A topological space Y is Hausdorff if, for any two distinct points y1, y2 in Y , there are disjoint open sets G1,G2 Y with y1 G1 and y2 G2. It is not hard to show that any compact subset⊆ of a Hausdorff∈ space is∈ closed. If X and Y are topological spaces then a map θ : X Y is continuous if, for all open sets G Y , the set θ−1(G) is open in X. Taking→ complements, we see that θ is continuous⊆ if and only if, for all closed K Y , the set θ−1(K) ⊆ is closed in X. If Z is a compact subset of X and θ : X Y is continuous, then θ(Z) is compact. → A homeomorphism from X to Y is a bijection X Y which is continuous and has a continuous inverse. If there is a homeomorphism→ from X to Y , we say that X and Y are homeomorphic. If X and Y are homeomorphic topological spaces then all of their topological properties are identical. In particular, X is compact if and only if Y is compact. Recall that if X is a topological space and Z X, then the subspace topology on Z is defined by declaring the open sets of⊆ Z to be the sets G Z for G an open set of X. Then Z is a compact subset of X if and only if Z∩ is a compact topological space (in the subspace topology). Also, it is easy to see that a subspace of a Hausdorff space is Hausdorff. 2.1.1 Lemma. Let X and Y be topological spaces, and suppose that X is compact and Y is Hausdorff. (i). If θ : X Y is a continuous bijection, then θ is a homeomorphism → onto Y . In particular, Y is compact. (ii). If θ : X Y is a continuous injection, then θ(X) (with the subspace topology→ from Y ) is homeomorphic to X. In particular, θ(X) is a compact subset of Y . 17 Proof. (i) Let K be a closed subset of X. Since X is compact, K is compact. Since θ is continuous, θ(K) is compact. A compact subset of a Hausdorff space is closed, so θ(K) is closed. Hence θ−1 is continuous, which shows that θ is a homeomorphism. So Y is homeomorphic to the compact space X; so Y is compact. (ii) The subspace θ(X) of the Hausdorff space Y is Hausdorff. Let θ : X θ(X), x θ(x), which is a continuous bijection. By (i), θ is a homeomorphism→ and7→ θ(X) is compact. e e 2.2 Subbases and weak topologies 2.2.1 Definition. If and are two topologies on a set X then we say T1 T2 that 1 is weaker than 2 if 1 2. [We mightT also say thatT isTsmaller⊆ T , or coarser than ]. T1 T2 2.2.2 Definition. Let (X, ) be a topological space. We say that a collection of open sets is a subbaseT for if is the weakest topology containing S ⊆ T T T . If the topology is understood, we will also say that is a subbase for theS topological spaceT X. S 2.2.3 Remark. Suppose that is a topology on X and . It is not hard to see that the collectionT of unions of finite intersectionS ⊆s T of sets in forms a topology on X which is no larger than . [The empty set is theS union of zero sets, and X is the intersection of zeroT sets, so and X are in this collection.] From this, it follows that that following conditions∅ are equivalent: (i). is a subbase for ; S T (ii). every set in is a union of finite intersections of sets in ; T S (iii). a set G X is in if and only if for every x G, there exist finitely many sets⊆ S ,S ,...,ST such that ∈ 1 2 n ∈ S x S S S G. ∈ 1 ∩ 2 ∩···∩ n ⊆ By the equivalence of (i) and (ii), if X,Y are topological spaces and is a subbase for X, then a map f : Y X is continuous if and only if f −1S(S) is open for all S . → ∈ S 18 2.2.4 Proposition. Let X be a set, let I be an index set and suppose that X is a topological space and f : X X for each i I. The collection i i → i ∈ = f −1(G): i I,G is an open subset of X S { i ∈ i} is a subbase for a topology on X, and this is the weakest topology such that f is continuous for all i I. i ∈ Proof. Let be the collection of all unions of finite intersections of sets from . ThenT is a topology on X and is a subbase for by the previous S ′T S T remark. If is any topology on X such that fi : X Xi is continuous T −1 → ′ for all i I then by the definition of continuity, fi (G) for all open ∈ ′ ∈ T subsets G Xi, so . Since is the weakest topology containing ⊆ S ⊆′ T T S this shows that , so is the weakest topology such that each fi is continuous. T ⊆ T T This allows us to introduce the following terminology. 2.2.5 Definition. If X is a set, X is a topological space and f : X X i i → i for i I then the weakest topology on X such that fi is continuous for each i I∈is called the weak topology induced by the family f : i I . ∈ { i ∈ } 2.2.6 Proposition. Suppose that X is a topological space with the weak topology induced by a family of maps fi : i I where fi : X Xi and Xi is a topological space for each i I. { ∈ } → ∈ If Y is a topological space then a map g : Y X is continuous if and only if f g : Y X is continuous for all i I.→ i ◦ → i ∈ −1 Proof. The sets fi (G) for i I and G an open subset of Xi form a subbase for X, by Proposition 2.2.4. Hence,∈ by Remark 2.2.3, g is continuous g−1(f −1(G)) is open for all i I and open G X ⇐⇒ i ∈ ⊆ i (f g)−1(G) is open for all i I and open G X ⇐⇒ i ◦ ∈ ⊆ i f g is continuous for all i I. ⇐⇒ i ◦ ∈ 2.2.7 Lemma. Suppose that X is a topological space with the weak topology induced by a family of mappings f : X X . If Y X then the weak { i → i}i∈I ⊆ topology induced by the family fi Y : Y Xi i∈I is the subspace topology on Y . { | → } Proof. Let g = f for i I. Observe that g−1(G)= f −1(G) Y for G an i i|Y ∈ i i ∩ open subset of Xi. It is not hard to check that the collection of all sets of this form is a subbase for both the weak topology induced by gi i∈I and for the subspace topology on Y . Hence these topologies are equal.{ } 19 2.3 The product topology and Tychonoff’s theorem If is a collection of open subsets of X, let us say that an open cover is an -coverS if every set in the cover is in . S S 2.3.1 Theorem (Alexander’s subbase lemma). Let X be a topological space with a subbase . If every -cover of X has a finite subcover, then X is compact. S S Proof. Suppose that the hypothesis holds but that X is not compact. Then there is an open cover with no finite subcover; ordering such covers by in- clusion we can apply Zorn’s lemma [FA2.15] to find an open cover C of X without a finite subcover that is maximal among such covers. Note that C cannot cover X by hypothesis, so there is some x X which does not lie∩ S in any set in the collection C . On the other hand,∈ C does cover X so there is some G C with x∩ SG. Since G is open and is a subbase, by Remark 2.2.3 we∈ have\ S ∈ S x S S G ∈ 1 ∩···∩ n ⊆ for some S1,...,Sn . For i = 1,...,n we have x Si so Si C. By the maximality of C, there∈ S is a finite subcover C of C S∈ . Let D6∈= C S . i ∪{ i} i i \{ i} Then D = D1 Dn covers X (S1 Sn).